1.6 Page 91: #1, 3, 5, 7, 13, 15, 17, 29-35 odd.

Homework:

Day 5: July 6thChapter: 1.5 Existence and Uniqueness.

Chapter: 1.6 Equilibria and Phase Line.

Coming up

Midterm 1 on Friday, July 9th: Chapter 1

Lab 1 due Thursday next week,July 15th

Existence and Uniquenessdy

dt= F (y, t)

Then: There exists a unique solution to initial value problem y(t0) = y0

defined for t0 !A < t < t0 + A

If: Continuously differentiable in and , i.e.ty

is differentiable in and and the derivative is continuousF y t

Existence and Uniquenessdy

dt= F (y, t)

Then: There exists a unique solution to initial value problem y(t0) = y0

defined for t0 !A < t < t0 + A

If: Continuously differentiable in and , i.e.ty

Result: Solutions can not cross.

Example: Homework. Chapter 1.5: 1,3,5 & 7.

Example: Homework. Chapter 1.5: 1,3,5 & 7.

1. dy

dt= f(t, y)

3. dy

dt= f(t, y)

y1(t) = 3 for all t is a solution

for all t is a solution

for all t is a solution

y1(t) = t + 2

what can we say about y(0) = 1

what can we say about y(0) = 1y2(t) = !t2

Example: Homework. Chapter 1.5: 1,3,5 & 7.

5.

7.

dy

dt= (y ! 2)(y ! 3)y

what can we say about

what can we say about y(0) = 1

y(0) = 4

E & U: Examples

dy

dt=

y

t

dy

dt=

y

t2

dy

dt=

32y

13

When existence or uniqueness fails

Look for “bad” points

1.

2.

3.

E & U: Examplesdy

dt=

y

t

Lots of solutions satisfy: y(0) = 0

but none satisfy y(0) = k != 0

1.

E & U: Examplesdy

dt=

y

t22. Homework: 1.5 #13

y(t) = 0 is an equilibrium solution

E & U: Examplesdy

dt=

y

t2

y(t) = De!1t

y(t)! 0t! 0+,As

t! 0!, y(t)! ±"

2. Homework: 1.5 #13

y(t) = 0 is an equilibrium solution

E & U: Examples

3. derivative not continuous at 0

y(t) = 0 is an equilibrium solution

dy

dt=

32y

13

Lots of solutions satisfy: y(0) = 0

Phase LineAutonomous equations

dy

dt= F (y)

slopes do not depend on t

sign of shows whether solutions increase or decrease.

F

Ex: Logistic equation

dy

dt= y(1! y)